(a) The sum of the digits of a two-digit number is 15. If the number formed by reversing the digits is less than the original number by 27, find the original number. |
(b) Verify that \[x=2\]is a solution of the equation |
\[2\left( x+1 \right)=3\left( x+1 \right)-3.\] |
Answer:
(a) Let the unit place = x Then the tens place \[=\left( 15x \right)\] Therefore, original number\[=10\left( 15x \right)+x\] \[=\left( 1509x \right)\] By reversing the digits, we get New number \[=10x+(15-x)\] \[=\text{ }9x\text{ }+\text{ }15\] According to question, (original number) - (new number) = 27 \[\left( 1509x \right)\left( 9x+15 \right)=27\] or \[13518x=27\] or \[18x=13527\] or \[18x=108\] or \[x\text{ }=\text{ }\frac{108}{18}\] or x = 6 Hence, original number \[=1509x\] \[=1509\times 6\] \[=15054=96\] (b) Verification: Since \[2\left( x+1 \right)=3\left( x+1 \right)\text{3}\] The putting x = 2, then LHS \[=\text{ }2\left( x\text{ }+\text{ }1 \right)\] = 2 (2 + 1) \[=2\times 3=6\] and RHS \[=3\left( x+1 \right)3\] \[=3\left( 2+1 \right)3\] \[=3\times 33\] \[=93=6\] Hence, LHS = RHS = 6 [Hence, Verified]
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